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AH
Akai Haruma
Giáo viên
10 tháng 8 2021

Lời giải:
a.

$7-3a=(\sqrt{7}-\sqrt{3a})(\sqrt{7}+\sqrt{3a})$

b. 

$14x^2-11=(\sqrt{14}x-\sqrt{11})(\sqrt{14}x+\sqrt{11})$

c.

$3x-6\sqrt{x}-6=3(x-2\sqrt{x}-2)$
$=3[(\sqrt{x}-1)^2-3]$

$=3(\sqrt{x}-1-\sqrt{3})(\sqrt{x}-1+\sqrt{3})$

d.

$x\sqrt{x}-3\sqrt{x}-2=x\sqrt{x}-2x+2x-4\sqrt{x}+\sqrt{x}-2$
$=x(\sqrt{x}-2)+2\sqrt{x}(\sqrt{x}-2)+(\sqrt{x}-2)$

$=(\sqrt{x}-2)(x+2\sqrt{x}+1)$

$=(\sqrt{x}-2)(\sqrt{x}+1)^2$

21 tháng 7 2016

a, \(x-\sqrt{x}\)\(\sqrt{x}.\left(\sqrt{x}-1\right)\)

b, 3x+6\(\sqrt{x}\)\(\sqrt{x}.\left(3\sqrt{x}+6\right)\)

c, x+2\(\sqrt{x}+1\)\(\left(\sqrt{x}\right)^2+2\sqrt{x}+1=\left(\sqrt{x}+1\right)^2\)

d, \(3x-5\sqrt{x}+2=3x-3\sqrt{x}-2\sqrt{x}+2\)

=\(3\sqrt{x}.\left(\sqrt{x}-1\right)-2.\left(\sqrt{x}-1\right)\)

=\(\left(3\sqrt{x}-2\right).\left(\sqrt{x}-1\right)\)

22 tháng 8 2015

2) a) \(x^2-3=\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\)

b) \(x^2-6=\left(x-\sqrt{6}\right).\left(x+\sqrt{6}\right)\)

c) = \(x^2+2x.\sqrt{3}+\left(\sqrt{3}\right)^2=\left(x+\sqrt{3}\right)^2\)

d) = \(x^2-2x\sqrt{5}+\left(\sqrt{5}\right)^2=\left(x-\sqrt{5}\right)^2\)

20 tháng 6 2021

a) 2a−4b=2(a−2b)2a−4b=2(a−2b)

c) 2ax−2ay+2a=2a(x−y+1)2ax−2ay+2a=2a(x−y+1)

e) 3xy(x−4)−9x(4−x)=3x(x−4)(y+3)3xy(x−4)−9x(4−x)=3x(x−4)(y+3)

b,d xem lại đề

20 tháng 6 2021

không hiểu

 what are you doing?

\(A,ĐKXĐ:x;y\ge0\)

\(A=\sqrt{xy}-2\sqrt{y}-5\sqrt{x}+10\)

\(=\sqrt{y}\left(\sqrt{x}-2\right)-5\left(\sqrt{x}-2\right)\)

\(=\left(\sqrt{x}-2\right)\left(\sqrt{y}-5\right)\)

\(ĐKXĐ:x;y\ge0\)

\(B=a\sqrt{x}+b\sqrt{y}-\sqrt{xy}-ab\)

\(=\left(a\sqrt{x}-\sqrt{xy}\right)+\left(b\sqrt{y}-ab\right)\)

\(=\sqrt{x}\left(a-\sqrt{y}\right)+b\left(\sqrt{y}-a\right)\)

\(=\sqrt{x}\left(a-\sqrt{y}\right)-b\left(a-\sqrt{y}\right)\)

\(=\sqrt{x}\left(a-\sqrt{y}\right)-b\left(a-\sqrt{y}\right)\)

\(=\left(a-\sqrt{y}\right)\left(\sqrt{x}-b\right)\)

19 tháng 7 2018

a ) \(x+\sqrt{x}=\left(\sqrt{x}\right)^2+\sqrt{x}=\sqrt{x}\left(\sqrt{x}+1\right)\)

b ) \(x-4\sqrt{x}+3=\left(\sqrt{x}\right)^2-2.\sqrt{x}.2+2^2-1=\left(\sqrt{x}-2\right)^2-1\)

\(=\left(\sqrt{x}-2\right)^2-1^2=\left(\sqrt{x}-2+1\right)\left(\sqrt{x}-2-1\right)=\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)\)

19 tháng 7 2018

\(x+\sqrt{x}=\left(\sqrt{x}\right)^2+\sqrt{x}=\sqrt{x}.\left(\sqrt{x}+1\right)\)

\(x-4\sqrt{x}+3=\left[\left(\sqrt{x}\right)^2-2.\sqrt{x}.2+2^2\right]-1^2=\left(\sqrt{x}-2\right)^2-1^2\)

\(=\left(\sqrt{x}-2-1\right)\left(\sqrt{x}-2+1\right)\)

\(=\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)\)